THE PICTURE IS BELOW DOES ANYBODY KNOW THIS!!

Mathematics

1 answer:

earnstyle [38]4 years ago

4 0

So I dont have all the answers but I hope this helps:

a) (5,4) and (6,3)

b) (1,3), (2,4), (5,7), (6,8)

Im stuck in the last question, but I hope this helped you

You might be interested in

What is the solution to -55+q=7 A. a=-62 B. q=62 C. q=42 D. q=-42

Scorpion4ik [409]

Answer:

C, as q = 62.

Step-by-step explanation:

When you have an equation, your goal is to get the letter you are solving for alone. To do this, you employ a simple rule: what you do to one side of the equals sign, you must do the other.

To isolate q in -55 + q = 7, you must add 55 to the left side. q is now alone. However, because we added 55 to the left side, we must also do it to the right! 7 + 55 = 62, so the new right side is 62. Hence, we get to this:

q = 62

The answer is now in plain sight!

4 0

3 years ago

Read 2 more answers

Which ordered pair is a reflection of (3, 7) across the y-axis?

stich3 [128]

If its a reflection across the y axis the x value is reflected and the y value stays the same, so (-3,7)

4 0

4 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$